Numerical peak holomorphic functions on Banach spaces - Core

J. Math. Anal. Appl. 364 (2010) 437–452

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Numerical peak holomorphic functions on Banach spaces Sung Guen Kim a,1 , Han Ju Lee b,∗,2 a b

Department of Mathematics, Kyungpook National University, Daegu 702-701, Republic of Korea Department of Mathematics Education, Dongguk University, Seoul 100-715, Republic of Korea

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 19 March 2009 Available online 27 October 2009 Submitted by Richard M. Aron Keywords: Numerical peak holomorphic functions Numerical boundary Numerical Shilov boundary

We introduce the notion of numerical (strong) peak function and investigate the denseness of the norm and numerical peak functions on complex Banach spaces. Let A b ( B X : X ) be the Banach space of all bounded continuous functions f on the unit ball B X of a Banach space X and their restrictions f | B ◦X to the open unit ball are holomorphic. In finite dimensional spaces, we show that the intersection of the set of all norm peak functions and the set of all numerical peak functions is a dense G δ -subset of A b ( B X : X ). We also prove that if X is a smooth Banach space with the Radon–Nikodým property, then the set of all numerical strong peak functions is dense in A b ( B X : X ). In particular, when X = L p (μ) (1 < p < ∞) or X = 1 , it is shown that the intersection of the set of all norm strong peak functions and the set of all numerical strong peak functions is a dense G δ -subset of A b ( B X : X ). As an application, the existence and properties of numerical boundary of A b ( B X : X ) are studied. Finally, the numerical peak function in A b ( B X : X ) is characterized when X = C ( K ) and some negative results on the denseness of numerical (strong) peak holomorphic functions are given. © 2009 Elsevier Inc. All rights reserved.

1. Introduction and preliminaries In this paper, we consider only complex Banach spaces. Given a Banach space X , we denote by B X and S X its closed unit ball and unit sphere, respectively. Let X ∗ be the dual space of X . If X and Y are Banach spaces, an N-homogeneous polynomial P from X to Y is a mapping such that there is an N-linear (bounded) mapping L from X to Y such that P (x) = L (x, . . . , x) for every x in X . P ( N X : Y ) denote the Banach space of all N-homogeneous polynomials from X to Y , endowed with the polynomial norm P = supx∈ B X P (x). A mapping Q : X → Y is a polynomial if there exist m and P k ∈ P (k X : Y ), k = 0, 1, . . . , m, such that Q = P 0 + P 1 + · · · + P m . If P m = 0, then we say that Q is a polynomial of degree m. We denote P ( X : Y ) the normed space of all polynomials from X to Y , endowed with the norm Q = supx∈ B X Q (x). We refer to [12] for background on polynomials. We are mainly interested in the following spaces. For two Banach spaces X , Y and a Hausdorff topological space K ,

C b ( K : Y ) := { f : K → Y : f is a bounded continuous function on K },

A b ( B X : Y ) := f ∈ C b ( B X : Y ): f is holomorphic on B ◦X ,

*

Corresponding author. E-mail addresses: [email protected] (S.G. Kim), [email protected] (H.J. Lee). 1 The author would like to thank the financial support of Kyungpook National University Research Fund, 2009. 2 The author was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2009-0069100). 0022-247X/$ – see front matter doi:10.1016/j.jmaa.2009.10.046

©

2009 Elsevier Inc. All rights reserved.

438

S.G. Kim, H.J. Lee / J. Math. Anal. Appl. 364 (2010) 437–452

A u ( B X : Y ) := f ∈ A b ( B X : Y ): f is uniformly continuous , where B ◦X is the interior of B X . Then C b ( K : Y ) is a Banach space under the sup norm f := sup{ f (t )Y : t ∈ K } and both A b ( B X : Y ) and A u ( B X : Y ) are closed subspaces of C b ( B X : Y ). In case that Y is the complex scalar field C, we just write C b ( B X ), A b ( B X ) and A u ( B X ). The closed subspace of A u ( B X : Y ) consisting of all weakly uniformly continuous functions is denoted by A wu ( B X : Y ). We denote by A ( B X : X ) one of A b ( B X : X ), A u ( B X : X ) and A wu ( B X : X ). Notice that if X is finite dimensional, A b ( B X : X ) = A u ( B X : X ) = A wu ( B X : X ). We denote by τ the product topology of the set S X × S X ∗ , where the topologies on S X and S X ∗ are the norm topology of X and the weak-∗ topology of X ∗ , respectively. The set Π( X ) := {(x, x∗ ): x = x∗ = 1 = x∗ (x)} is a τ -closed subset of S X × S X ∗ . The spatial numerical range of f in C b ( B X : X ) is defined by W ( f ) = {x∗ ( f (x)): (x, x∗ ) ∈ Π( X )}, and the numerical radius of f is defined by v ( f ) = sup{|λ|: λ ∈ W ( f )}. Let f be an element of C b ( K : X ). We say that f attains its norm if there is some t ∈ K such that f = f (t ) X . f is said to be a (norm) peak function at t if there exists a unique t ∈ K such that f = f (t ) X . It is clear that every (norm) peak function in C b ( K : X ) is norm attaining. A peak function f at t is said to be a (norm) strong peak function if whenever there is a sequence {tk }k∞=1 in K with limk f (tk ) X = f , {tk }k∞=1 converges to t. It is easy to see that if K is compact, then every peak function is a strong peak function. Given a subspace H of C b ( K ), we denote by ρ H the set of all points t ∈ K such that there is a strong peak function f in H with f = | f (t )|. Recently, the first named author [15] introduced and studied the concepts of numerical peak functions and numerical peak points. Let f be an element of C b ( B X : X ), where X is a Banach space. If there is some (x, x∗ ) in Π( X ) such that v ( f ) = |x∗ ( f (x))|, we say that f attains its numerical radius. f is said to be a numerical peak function at (x, x∗ ) if there exists a unique (x, x∗ ) ∈ Π( X ) such that v ( f ) = |x∗ ( f (x))|. In this case, (x, x∗ ) is said to be the numerical peak point of f . The numerical peak function f at (x, x∗ ) is called a numerical strong peak function if whenever there is a sequence {(xk , xk∗ )}k∞=1 in Π( X ) such that limk |xk∗ ( f (xk ))| = v ( f ), then {(xk , xk∗ )}k∞=1 converges to (x, x∗ ) in τ -topology. In this case, (x, x∗ ) is said to be the numerical strong peak point of f . We say that a numerical strong peak function f at (x, x∗ ) is said to be a very strong numerical peak function if whenever there is a sequence {(xk , xk∗ )}k∞=1 in Π( X ) satisfying limn |xk∗ ( f (xk ))| = v ( f ), we get limk xk = x and limk xk∗ = x∗ in the norm topology. If X is finite dimensional, then every numerical peak function is a very strong numerical peak function. A function f in C b ( B X : K ) is called a norm and numerical (resp. strong) peak function if it is both a norm (resp. strong) and a numerical (resp. strong) peak function. In 1996, Y.S. Choi and the first named author [9] initiated the study of denseness of norm or numerical radius attaining nonlinear functions, especially, homogeneous polynomials on a Banach space. Using the perturbed optimization theorem of Bourgain [6] and Stegall [19], they proved that if X has the Radon–Nikodým property, then the set of all norm attaining functions in P (k X ) is norm-dense. Concerning the numerical radius, it was also shown that if X has the Radon–Nikodým property, then the set of all numerical radius attaining functions in P (k X : X ) is norm-dense. M.D. Acosta, J. Alaminos, D. García and M. Maestre [1] proved that if X has the Radon–Nikodým property, then the set of all norm attaining functions in A b ( B X ) is norm-dense. Recently, it was shown [10] that if X has the Radon–Nikodým property, the set of all (norm) strong peak functions in A b ( B X ) is dense. Concerning the numerical radius, M.D. Acosta and the first named author [2] showed that the set of all numerical radius attaining functions in A b ( B X : X ) is dense if X has the Radon–Nikodým property. In this paper, we extend the results of the above [9,1,10,2] to the denseness of norm or numerical (strong) peak functions in A ( B X : X ) if X has the Radon–Nikodým property. Let’s briefly sketch the content of this paper. In Section 2, we show that if X is a finite dimensional Banach space, then the set of all norm and numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). Let K be a convex subset of a Banach space X . An element x in K is said to be a strongly exposed point of K if there is nonzero x∗ ∈ B X ∗ such that Re x∗ (x) = sup{Re x∗ ( y ): y ∈ K } and whenever limn Re x∗ (xn ) = Re x∗ (x) for some sequence {xn }n∞=1 in K , we get limn xn − x = 0. A Banach space X is said to have the Radon–Nikodým property if every nonempty bounded closed convex subset in X is a closed convex hull of its strongly exposed points [11]. The point x ∈ B X is said to be a smooth point if there is a unique x∗ ∈ B X ∗ such that Re x∗ (x) = 1. We denote by sm( B X ) the set of all smooth points of B X . We say that a Banach space is smooth if sm( B X ) is the unit sphere S X . When X is a smooth Banach space with the Radon–Nikodým property, it is shown that the set of all numerical strong peak functions is dense in A ( B X : X ). Moreover, if X is a Banach space with the Radon–Nikodým property and X ∗ is locally uniformly convex, then the set of all norm and numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). As a corollary, if 1 < p < ∞ and X = L p (μ) for a measure space μ, then the set of all norm and numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). In this case, every numerical strong peak function is a very strong numerical peak function. We also prove that the set of all norm and numerical strong peak functions in A ( B l1 : l1 ) is a dense G δ -subset of A ( B l1 : l1 ). We also give some applications of the denseness of numerical strong peak holomorphic functions. M.D. Acosta and the first named author [3,15] introduced and studied the concepts of numerical boundary and numerical Shilov boundary for a subspace H of C b ( B X : X ). A subset Γ of Π( X ) is called a numerical boundary for a subspace H of C b ( B X : X ) if v ( f ) = sup{|x∗ ( f (x))|: (x, x∗ ) ∈ Γ } for every f in H . If it exists, the smallest closed numerical boundary for a subspace H , it is called the numerical Shilov boundary of H . We show that if the set of numerical strong peak functions are dense in a subspace A of C b ( B X : X ) then the numerical Shilov boundary of A exists and it is the τ closure of the set of all numerical strong peak points. Recently, it is shown [16] that if Π( X ) is metrizable and the set Γ = {(x, x∗ ) ∈ Π( X ): x ∈ ρ A ( B X ) ∩ sm( B X )} is a numerical boundary of A ( B X : X ), then the set of all numerical strong peak functions is dense in A ( B X : X ).


439

In Section 3, we characterize the numerical peak functions in A b ( B X : X ) when X = C ( K ), where K is a compact Hausdorff space. More precisely, setting X = C ( K ), an element f in A b ( B X : X ) is a numerical peak function in A b ( B X : X ) if and only if there exist unique x0 ∈ ext( B C ( K ) ) and t 0 ∈ K such that (a) v ( f ) = f = f (x0 ) > f (x) for every x ∈ B C ( K ) with x = x0 ; (b) v ( f ) = f = |δt0 ( f (x0 ))| > |δt ( f (x0 ))| for every t ∈ K with t = t 0 . For negative results for denseness of numerical peak holomorphic functions on a classical Banach space, we prove the following: (1) Let Ω be a locally compact Hausdorff space with more than 2 elements. Let X = C 0 (Ω). Then there are no numerical peak functions in A b ( B X : X ). (2) Let K be an infinite compact Hausdorff space. Then there are no numerical strong peak functions in A b ( B C ( K ) : C ( K )). Neither are there in A wu ( B L 1 [0,1] : L 1 [0, 1]). 2. Denseness of numerical peak holomorphic functions Let’s begin with the basic properties of the set of strong peak functions and the set of numerical strong peak functions. Recall that a Banach space X is said to be locally uniformly convex if for every x ∈ S X and every sequence {xn } in B X satisfying limn xn + x = 2, we get limn xn − x = 0. Proposition 2.1. Let X , Y be Banach spaces. Let A be a subspace of C b ( B X : Y ). Then the set of all strong peak functions in A is a G δ -subset of A. In case that Π( X ) is a complete metrizable space and X = Y , then the set of all numerical strong peak functions in A is a G δ -subset of A. If, moreover, X ∗ is locally uniformly convex, then Π( X ) is complete metrizable and the net convergence of {(xα , x∗α )}α in τ -topology implies the convergence of each component in norm. Proof. Notice that, since B X is metrizable and complete, f ∈ A is a strong peak function at x0 if and only if for each neighborhood V of x0 , there is > 0 such that {x ∈ B X : f (x) f − } is contained in V . For each f ∈ A and n 1, let

1 . ( f , n) = x ∈ B X : f (x) f − n

For each natural number N 1, define

S N :=

∞

f ∈ A \ {0}: diam ( f , n)

n =1

1

,

N

where ∞ diam(C ) = sup{d(x, y ): x, y ∈ A } for a metric space (C , d). Notice that the set of all strong peak functions is S := N =1 S N . So we have only to show that each S N is an open subset of A. Fix f ∈ S N , then there is n 1 such that n f > 1 1 B A is a subset of S N . and diam ( f , n) N1 . If g − f 1/(3n), then g = 0 and ( g , 3n) ⊂ ( f , n). This shows that f + 3n Hence S N is an open subset of A. For the second case, notice that, since Π( X ) is metrizable and complete, f ∈ A is a numerical strong peak function at (x0 , x∗0 ) if and only if for each τ -neighborhood V of (x0 , x∗0 ), there is > 0 such that {(x, x∗ ) ∈ Π( X ): |x∗ ( f (x))| v ( f ) − } is contained in V . For each f ∈ A and n 1, define

˜ f , n) = (

∗

x, x

∗

1 ∈ Π( X ): x f (x) v ( f ) − . n

For each natural number N 1, define

S N :=

∞

˜ f , n) f ∈ A \ {0}: diam (

n =1

1

N

. ∞

Notice that the set of all numerical strong peak functions is S := N =1 S N . So we have only to show that each S N is an ˜ f , n) 1 . If g − f 1/(3n), then g = 0 open subset of A. Fix f ∈ S N , then there is n 1 such that n f > 1 and diam ( N

˜ g , 3n) ⊂ ( ˜ f , n). This shows that f + 1 B A is a subset of S N . Hence S N is an open subset of A. and ( 3n Finally, suppose that X ∗ is locally uniformly convex. Then define a function d in Π( X ) × Π( X ) to be

d x, x∗ , y , y ∗

:= x − y + x∗ − y ∗ .

440


It is clear that d is a complete metric in Π( X ) and it is also clear that the d-convergence implies the τ -convergence. For the converse, if the net {(xα , x∗α )}α converges to ( z, z∗ ) in τ -topology, then limα z − xα = 0 and {x∗α }α converges weak-∗ to z∗ with z∗ ( z) = 1. So

∗ x∗α + z∗ ∗ lim sup xα + z 1. 1 lim inf α 2 2 α

Since X ∗ is locally uniformly convex, limα x∗α + z = 2 implies that limα z∗ − x∗α = 0. Hence the net {(xα , x∗α )} converges to ( z, z∗ ) in the d-metric topology. This completes the proof. 2 In case that X is finite dimensional, it is clear that Π( X ) is a compact metric space. For a separable Banach space, we get the following. Proposition 2.2. Let X be a separable Banach space. Then Π( X ) is complete and metrizable. Proof. Let {cn }n∞=1 be a dense subset in B X . Then in B X ∗ , the metric

d x∗ , y ∗ :=

∞ |x∗ (cn ) − y ∗ (cn )|

2n

n =1

induces the same topology as the weak-∗ topology in B X ∗ . Define a function d1 : Π( X ) × Π( X ) → [0, ∞) to be

d1 x, x∗ , y , y ∗

:= x − y + d x∗ , y ∗ .

It is clear that d1 induces the τ -topology in Π( X ). So we have only to show that d1 is a complete metric. Suppose that {(xn , xn∗ )} is a d1 -Cauchy sequence. Then it is clear that there is x ∈ S X such that limn xn − x = 0. Notice that limn xn∗ (ck ) exists for each k 1. Let x∗ be the weak-∗ limit point of {xn∗ }n∞=1 . Then x∗ (ck ) = limn xn∗ (ck ) for each k 1. Hence {xn∗ }n converges weak-∗ to x∗ . This completes the proof. 2 Combining Propositions 2.1 and 2.2, we get the following. Corollary 2.3. Let X be a separable Banach space and let A be a subspace of C b ( B X : X ). Then the set of all numerical strong peak functions in A is a G δ -subset of A. 2.1. Finite dimensional spaces Let K be a Hausdorff space and Y be a complex Banach space. Consider the product space K × B Y ∗ where B Y ∗ is equipped with the weak-∗ topology. Given a subspace A of C b ( K : Y ), consider the map ϕ : f ∈ A → ˜f ∈ C b ( K × B Y ∗ ) defined by

˜f x, y ∗ = y ∗ f (x) ,

∀ x, y ∗ ∈ K × B Y ∗ .

˜ of A is also a subspace of C b ( K × B Y ∗ ). We say that the subspace A of C b ( K : Y ) Then ϕ is a linear isometry, and its image A is separating if the following conditions hold: ˜ for every x∗ , y ∗ ∈ S Y ∗ , (i) if x = y in K , then δ(x,x∗ ) = δ( y , y ∗ ) on A

˜ for every x∗ = y ∗ in ext( B Y ∗ ), (ii) given x ∈ K with δx = 0 on A, we have δ(x,x∗ ) = δ(x, y ∗ ) on A where δt (for some t ∈ K ) is a linear map from C b ( K : Y ) defined by δt ( f ) = f (t ) and the ext(C ) is the set of all extreme points of a convex set C . We need the following theorem from [10]. Theorem 2.4. (See [10].) Let Y be a Banach space and let A be a nontrivial separating separable subspace of C ( K : Y ) on a compact Hausdorff space K . Then the set { f ∈ A: f is a peak function at some t ∈ K , f (t )/ f ∈ sm( B Y )} is a dense G δ -subset of A. Let H be a subspace of C b ( B X : X ) and define H -numerical index N H ( X ) by N H ( X ) = inf{ v ( f ): f = 1, f ∈ H }. The H -numerical index of X is a direct generalization of the k-polynomial numerical index which is the constant N H ( X ) when H = P (k X : X ) [7]. Proposition 2.5. Let X be a finite dimensional Banach space. Suppose that a subspace H of C b ( B X : X ) contains the functions of the form

1 ⊗ x,

y ∗ ⊗ z,

∀x, z ∈ X , ∀ y ∗ ∈ X ∗ .

If the H -numerical index N H ( X ) > 0, then the set of all numerical peak functions in H is a dense G δ -subset of H .

(2.1)


441

Proof. Suppose that N H ( X ) > 0. Then N H ( X ) f v ( f ) f for all f ∈ H . So the v (·) is a complete norm on H . Consider the linear map f → ˆf from H into C (Π( X )) defined by

ˆf x, x∗ = x∗ f (x) .

ˆ be the image of H in C (Π( X )). So the two Banach spaces ( H , v ) and ( Hˆ , · ) Notice that v ( f ) = ˆf for every f ∈ H . Let H are isometrically isomorphic. ˆ is a separable subspace of Since X is finite dimensional, Π( X ) is compact metrizable so C (Π( X )) is separable. Then H ˆ is separating in C (Π( X )) and using Theorem 2.4. C (Π( X )) and we are done by proving that H ˆ is separating. Claim: H Let (x, x∗ ) = ( y , y ∗ ) ∈ Π( X ) and let 1 ⊗ z ∈ H . Then

α , β ∈ S C . If α x∗ = β y ∗ , then choose z ∈ S X such that α x∗ (z) = β y ∗ (z). Set f :=

α δ(x,x∗ ) ( ˆf ) = α ˆf x, x∗ = α x∗ (z) = β y ∗ (z) = β ˆf y , y ∗ = βδ( y, y∗ ) ( ˆf ). Now suppose that α = 0 and

α x∗ = β y ∗ . Then x = y, and choose z∗ ∈ S X ∗ such that z∗ (x) = z∗ ( y ). Set g := z∗ ⊗ x ∈ H . Then β y ∗ (x) =

α gˆ x, x∗ = α z∗ (x)x∗ (x) = β z∗ (x) y ∗ (x) = β z∗ ( y ) y ∗ (x) = β gˆ y , y ∗ . Hence

α δ(t ,t ∗ ) ( gˆ ) = βδ(s,s∗ ) ( gˆ ). Therefore Hˆ is a separating separable subspace of C (Π( X )). By Theorem 2.4, the set of peak

ˆ is dense. So we get the desired result. functions in H

2

Recall the following theorem of L.A. Harris [14]. Theorem 2.6 (Harris). Let h ∈ A b ( B X : X ) and P m the mth term of the Taylor series expansion for h about 0. Then P m km v (h), where k0 = 1, k1 = e and km = mm/(m−1) for m 2. Proposition 2.7. Let m 1 be a natural number and H m be the subspace of A b ( B X : X ) consisting of all polynomials of degree m. Then its numerical index N H m ( X ) is positive. Proof. Let h ∈ H m and x ∈ B ◦X . Then h(x) = m

m

k =0

k =0

P k (x)

where cm =

m

k=0 km

Pk

m

m

k=0

P k (x). Then by Theorem 2.6,

km v (h) cm v (h),

k =0

−1 > 0. 2 > 0. Hence h cm v (h). Therefore N H m ( X ) cm

From Propositions 2.5 and 2.7, we have the following. Proposition 2.8. Let m 1 be a natural number and H m be the subspace of A b ( B X : X ) consisting of all polynomials of degree m. If X is finite dimensional, the set of all numerical peak functions in H m is a dense G δ -subset of H m . Theorem 2.9. Let X be a finite dimensional complex Banach space. Then the set of all norm and numerical peak functions in A u ( B X : X ) is dense. In fact, setting

1 = f ∈ A u ( B X : X ): f is peak function at t ∈ B X and f (t )/ f is a smooth point of B X , 2 = f ∈ A u ( B X : X ): f is a numerical peak function , the intersection 1 ∩ 2 is a dense G δ -subset of A u ( B X : X ). Proof. Notice that if X is a finite dimensional Banach space, then the subspace H m ⊂ A u ( B X : X ) of all polynomials of degree m for some m 1 is a separating subspace of C ( B X : X ). Hence, by Theorem 2.4 and Proposition 2.8, the intersection 1 ∩ 2 ∩ H m is a dense G δ -subset of H m . So we get the following theorem. Let > 0 and f ∈ A u ( B X : X ). Then choose pm be a polynomial of degree m such that f − pm . So there is a norm and numerical peak function q ∈ 1 ∩ 2 ∩ H m such that q − pm . Hence f − q 2 . So Proposition 2.1 completes the proof. 2 For later use, we end this section with a basic property of numerical radius of holomorphic function.

442


Lemma 2.10. Let X be a Banach space and f ∈ A b ( B X : X ). Suppose that there is y in B X and y ∗ ∈ B X ∗ such that | y ∗ ( y )| = y ∗ · y . Then | y ∗ ( f ( y ))| v ( f ). In particular, f (0) v ( f ). Proof. If y ∗ = 0, then it is clear. So we may assume that y ∗ = 0. Suppose first that y = 0. By the Bishop–Phelps theorem [5], given > 0, there is w ∗ ∈ B X ∗ \ {0} such that w ∗ − y ∗ and w ∗ attains its norm at some x ∈ S X . Then by the maximum modulus theorem,

∗ ∗

y f (0) w ∗ f (0) + f (0) w w ∗ f (0) + f (0) ∗ w

max ∗ f (λx) + f (0) v ( f ) + f (0). |λ|=1 w

Since > 0 is arbitrary, | y ∗ ( f (0))| v ( f ). In case that y = 0, then again by the maximum modulus theorem,

∗ y ∗

∗

y v ( f ). y f ( y) y max ∗ f λ y ∗ f ( y ) |λ|= 1 y y

This completes the proof.

2

2.2. Banach spaces with the RNP An element h ∈ A b ( B X : X ) is said to strongly attain its numerical radius if there is (x, x∗ ) ∈ Π( X ) such that whenever there is a sequence {(xn , xn∗ )}n∞=1 in Π( X ) with limn |xn∗ (h(xn ))| = v (h), there exist a subsequence {(xnk , xn∗k )}k∞=1 in Π( X ) and

λ ∈ S C such that {(xnk , xn∗k )}k∞=1 converges to (λx, λx∗ ) in Π( X ).

Lemma 2.11. Let X be a smooth Banach space with the Radon–Nikodým property and let N 1 be a natural number. Then for each > 0 and for each f ∈ A ( B X : X ), there is a Q ∈ P (N X : X ) such that Q < and f + Q strongly attains its numerical radius. In particular, the set of all strongly numerical radius attaining elements of A ( B X : X ) (resp. P ( N X : X )) is dense in A ( B X : X ) (resp. P (N X : X )). Proof. Fix f ∈ A b ( B X : X ) and

> 0. Define for each x ∈ B X , ∗

ϕ (x) := max x f (λx) : λ ∈ C, |λ| 1, x∗ (x) = x, x∗ ∈ S X .

We claim that ϕ is upper semi-continuous. Indeed, if the sequence {xn }n∞=1 converges to x, then for each n 1, choose λn and xn∗ such that ϕ (xn ) = |xn∗ ( f (λn xn ))| and let x∗ be the weak-∗ limit point of {xn∗ }. Then since xn∗ (xn ) = xn , we get x∗ (x) = x. We may assume that the sequence {xn }n∞=1 and {λn }n∞=1 converge to x∗ and λ, respectively. Then

lim

n→∞

ϕ (xn ) = limxn∗ f (λn xn ) = x∗ f (λx) ϕ (x). n

Hence it is easy to see that lim supn ϕ (xn ) ϕ (x). By the perturbed optimization theorem of Bourgain and Stegall [6,19], there is y ∗ such that y ∗ < and strongly exposes B X at x0 . Then y ∗ (x0 ) = 0. Otherwise,

ϕ (x0 ) = sup ϕ (x) + Re y ∗ (x): x ∈ B X

ϕ + Re y ∗

= sup ϕ (x) + y ∗ (x): x ∈ B X

and ϕ (x0 ) + Re y ∗ (x0 ) = ϕ (−x0 ) + Re y ∗ (−x0 ). Since ϕ + Re y ∗ strongly exposes B X at x0 , we get x0 = 0. It is clear that ϕ (0) = supx∈ B X ϕ (x) v ( f ). This implies that ϕ (0) = f (0) = v ( f ) by Lemma 2.11. Choose a sequence {(xn , xn∗ )}n∞=1 in Π( X ) such that

limxn∗ f (xn ) = v ( f ). n

Then |xn∗ ( f (xn ))| ϕ (xn ) + | y ∗ (xn )| = ϕ (λn xn ) + Re y ∗ (λn xn ) ϕ (0) = v ( f ) for a suitable sequence {λn } in S C . So limn ϕ (λn xn ) + Re y ∗ (λn xn ) = ϕ (0). Since ϕ + Re y ∗ strongly exposes B X at 0, {λn xn }n∞=1 converges to 0. This is a contradiction to limn λn xn = 1. Now we get x0 = 1. Indeed, it is clear that x0 = 0. If 0 < x0 < 1, then

ϕ (x0 ) + Re y ∗ (x0 ) = sup ϕ (x) + Re y ∗ (x): x ∈ B X

= sup ϕ (x) + y ∗ (x): x ∈ B X


443

shows that Re y ∗ (x0 ) = | y ∗ (x0 )| and

ϕ (x0 ) + y ∗ (x0 ) < ϕ

x0 x0 ∗ = ϕ x0 + y ∗ + . Re y x0 x0 x0 x0 x0

This is a contradiction to the fact that ϕ + Re y ∗ strongly exposes B X at x0 . There exist λ0 ∈ S C , x∗0 ∈ S X ∗ , and x∗0 (x0 ) = 1 such that ϕ (x0 ) = |x∗0 ( f (λ0 x0 ))|. Define h : B X → X by

N −1

h(x) := f (x) + λ1 λ0 x∗0 (x)

y ∗ (x)x0 ,

where the complex number λ1 ∈ S C is properly chosen so that

∗

x f (λ0 x0 ) + λ1 λ0 y ∗ (x0 ) = x∗ f (λ0 x0 ) + y ∗ (x0 ). 0

0

It is clear that h ∈ A ( B X : X ) and notice that we get for every (x, x∗ ) ∈ Π( X ),

∗

x h(x) x∗ f (x) + y ∗ (x) ϕ (x) + y ∗ (x) sup ϕ (x) + y ∗ (x): x ∈ B X = sup ϕ (x) + Re y ∗ (x): x ∈ B X = ϕ (x0 ) + Re y ∗ (x0 ).

(2.2)

Note that (λ0 x0 , λ0 x∗0 ) ∈ Π( X ). Hence v (h) = ϕ (x0 ) + Re y ∗ (x0 ) because Re y ∗ (x0 ) = | y ∗ (x0 )| and

v (h) λ0 x∗0 h(λ0 x0 ) = x∗0 h(λ0 x0 ) = x∗0 f (λ0 x0 ) + λ1 λ0 y ∗ (x0 )

= x∗0 f (λ0 x0 ) + y ∗ (x0 ) = ϕ (x0 ) + y ∗ (x0 ) = ϕ (x0 ) + Re y ∗ (x0 ).

We shall show that h strongly attains its numerical radius at (x0 , x∗0 ). Suppose that limn |xn∗ (h(xn ))| = v (h) = ϕ (x0 ) + Re y ∗ (x0 ). Choose a sequence {αn } of complex numbers so that |αn | = 1 and

ϕ (xn ) + y ∗ (xn ) = ϕ (αn xn ) + Re y ∗ (αn xn ), ∀n 1. Then (2.2) shows that limn→∞ ϕ (αn xn ) + Re y ∗ (αn xn ) = ϕ (x0 ) + Re y ∗ (x0 ). Since ϕ + Re y ∗ strongly exposes B X at x0 , {αn xn } converges to x0 . Hence there is a subsequence of {xn } which converges to α x0 for some |α | = 1. For any weak-∗ limit point x∗ of {xn∗ }, x∗ (α x0 ) = 1. Since X is smooth, x∗ = α x∗0 . This shows that the subsequence {xn∗ } converges weak-∗ to α x∗0 . This shows that h strongly attains its numerical radius at (x0 , x∗0 ). Notice that h − f . This completes the proof. 2 Theorem 2.12. Suppose that X is a smooth Banach space with the Radon–Nikodým property. Then the set of all numerical strong peak functions is dense in A ( B X : X ). Proof. By the aid of Lemma 2.11, choose h ∈ A ( B X : X ) which strongly attains its numerical radius at (x0 , x∗0 ), h − f < 2 , and v (h) = |x∗0 (h(λ0 x0 ))| for some (x0 , x∗0 ) ∈ Π( X ) and λ0 ∈ S C . Choose a peak function g ∈ A ( B C ) at λ0 with g < 2 . Define

u (x) := h(x) + η g x∗0 (x) x0 , where

η ∈ S C is chosen to be |x∗0 (h(λ0 x0 )) + η g (λ0 )| = |x∗0 (h(λ0 x0 ))| + | g (λ0 )|. Then ∗

x u (λ0 x0 ) = x∗ h(λ0 x0 ) + η g (λ0 ) 0 0

= x∗0 h(λ0 x0 ) + g (λ0 ) = v (h) + g .

For each (x, x∗ ) ∈ Π( X ), we have

∗

x u (x) x∗ h(x) + g x∗ (x) v (h) + g . 0

So v (u ) = v (h) + g . Now we claim that u is a numerical strong peak function at (λ0 x0 , λ0 x∗0 ). If there is a sequence {(xn , xn∗ )} in Π( X ) with limn |xn∗ (u (xn ))| = v (u ), then

∗

x u (xn ) x∗ h(xn ) + g x∗ (xn ) v (h) + g . n

n

0

Hence limn |xn∗ (h(xn ))| = v (h) and limn | g (x∗0 (xn ))| = g . Since g is a strong peak function at λ0 , we get limn x∗0 (xn ) = λ0 . For any subsequence of {(xn , xn∗ )}n∞=1 , there are a further subsequence {( yn , yn∗ )} and η ∈ S C such that limn yn = η x0 and ∗ w − limn yn∗ = η x∗0 . Since limn x∗0 ( yn ) = η , η = λ0 . This implies that limn xn = λ0 x0 . Let x∗ be the weak-∗ limit point of {xn∗ }n∞=1 . Since X is smooth, x∗ (x0 ) = λ0 implies that x∗ = λ0 x∗0 . Therefore the weak-∗ limit of {xn∗ }n∞=1 is λ0 x∗0 . Thus u is a numerical strong peak function at (λ0 x0 , λ0 x∗0 ). Notice also that f − u . This completes the proof. 2 Question. Suppose that X is a smooth Banach space with the Radon–Nikodým property. Is it true that the set of all elements which are norm and numerical strong peak functions is dense in A ( B X : X )?

444


It is shown in [10] that the set of strong peak functions in A ( B X : X ) is dense if X has the Radon–Nikodým property. Hence by Theorem 2.12 and Proposition 2.1 we get the following. Corollary 2.13. Let X be a complex Banach space with the Radon–Nikodým property and X ∗ is locally uniformly convex. Then the set of all norm and numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). In particular, every numerical strong peak function is a very strong numerical peak function. Corollary 2.14. Let 1 < p < ∞ and X = L p (μ) for some measure space (Ω, Σ, μ). Then the set of all norm and numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). In particular, every numerical strong peak function is a very strong numerical peak function. The space 1 has the Radon–Nikodým property but it is not smooth. However, the following result analogous to Theorem 2.12 holds. Theorem 2.15. Let X = 1 . Then the set of all numerical strong peak functions in A ( B X : X ) is dense in A ( B X : X ). Proof. Fix f ∈ A ( B X : X ) and

> 0. Define for each x ∈ B X , ∗

ϕ (x) := max x f (λx) : λ ∈ C, |λ| 1, x∗ (x) = x, x∗ ∈ S X ∗ .

Then ϕ is upper semi-continuous by the similar argument in the proof of Lemma 2.11. Since 1 has the Radon–Nikodým property, there is y ∗ such that y ∗ < and ϕ + Re y ∗ strongly exposes B X at x0 . Then ∗ y (x0 ) = 0 and x0 = 1. On the contrary, if y ∗ (x0 ) = 0, then

ϕ (x0 ) = sup ϕ (x) + Re y ∗ (x): x ∈ B X

= sup ϕ (x) + y ∗ (x): x ∈ B X

and ϕ (x0 ) + Re y ∗ (x0 ) = ϕ (−x0 ) + Re y ∗ (−x0 ). Since ϕ + Re y ∗ strongly exposes B X at x0 , we get x0 = 0. It is clear that ϕ (0) = supx∈ B X ϕ (x) v ( f ). This implies that ϕ (0) = f (0) = v ( f ) by Lemma 2.11. Choose a sequence {(xn , xn∗ )}n∞=1 in Π( X ) such that

limxn∗ f (xn ) = v ( f ). n

Then |xn∗ ( f (xn ))| ϕ (xn ) + | y ∗ (xn )| = ϕ (λn xn ) + Re y ∗ (λn xn ) ϕ (0) = v ( f ) for a suitable sequence {λn } in S C . So limn ϕ (λn xn ) + Re y ∗ (λn xn ) = ϕ (0). Since ϕ + Re y ∗ strongly exposes B X at 0, {λn xn }n∞=1 converges to 0. This is a contradiction to limn λn xn = 1. Now x0 = 1. Indeed, it is clear that x0 = 0. If 0 < x0 < 1, then


= sup ϕ (x) + y ∗ (x): x ∈ B X

shows that Re y ∗ (x0 ) = | y ∗ (x0 )| and

ϕ (x0 ) + y ∗ (x0 ) < ϕ

∗ x0 x0 x0 ∗ + y =ϕ + Re y . x0 x0 x0 x0 x0

This is a contradiction to the fact that For later use, notice that

ϕ + Re y ∗ strongly exposes B X at x0 .


= sup ϕ (x) + y ∗ (x): x ∈ B X = ϕ (x0 ) + y ∗ (x0 ).

By the definition of

ϕ , there exist λ0 ∈ S C and x∗0 ∈ S ∞ such that ϕ (x0 ) = |x∗0 ( f (λ0 x0 ))| and x∗0 (x0 ) = 1 = x0 . So it is

easy to see that if i ∈ supp(x0 ), then x∗0 (i ) = sign(x0 (i )). For each i 1, let x∗1 as

x∗1 (i ) := Notice that

sign(x0 (i )), 0,

if x0 (i ) = 0, otherwise.


ϕ (x0 ) =

y ∗ ∈ B ∞

=

445

∗ x f (λ0 x0 ) + y ∗ f (λ0 x0 )

sup

, supp( y ∗ )∩supp(x∗ )=∅ 1

sup

y ∗ ∈ B ∞ , supp( y ∗ )∩supp(x∗1 )=∅

= x∗ f (λ0 x0 ) +

1

∗ x f (λ0 x0 ) + y ∗ f (λ0 x0 ) 1

∗ e , f (λ0 x0 ) , i

1

i∈ / supp(x0 )

where e ∗i is the ith coordinate functional defined by e ∗i (x) = x(i ) (x ∈ 1 ). Now let x∗2 as

x∗2 (i ) :=

sign( f (λ0 x0 ))(i ), 0,

if x0 (i ) = 0, i ∈ supp( f (λ0 x0 )), otherwise.

Then ϕ (x0 ) = |x∗1 ( f (λ0 x0 ))| + |x∗2 ( f (λ0 x0 ))|. Now we shall prove the theorem for three different cases according to the support of x0 and f (λ0 x0 ). In the first case, suppose that N = supp(x0 ) ∪ supp( f (λ0 x0 )) and

∗ e , f (λ0 x0 ) = 0. i

i∈ / supp(x0 )

Choose a peak function g ∈ A u ( B C ) at λ0 with 0 < g and 0 = y ∈ 1 such that supp( y ) is N \ (supp(x0 ) ∪ supp( f (λ0 x0 ))) and y . Then there is a unique element x∗3 ∈ S ∞ such that supp(x∗3 ) = supp( y ) and x∗3 ( y ) = y . Define a function h ∈ A ( B X : X ) as

h(x) := f (x) + η1 y ∗ (x)x0 + η2 g x∗1 (x) x0 + y , and define z∗

:=

x∗

1

+ ξ2

x∗

numbers in S C such that

2

+ ξ3

x∗

3

∈

∀x ∈ B X , ∗ S l∞ , where if x1 ( f (λ0 x0 )) = 0, then

η1 , η2 , ξ2 and ξ3 are uniquely determined complex

∗

z h(λ0 x0 ) = x∗ f (λ0 x0 ) + ξ2 x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ3 y 1 2 = x∗1 f (λ0 x0 ) + x∗2 f (λ0 x0 ) + y ∗ (λ0 x0 ) + g + y = ϕ (x0 ) + y ∗ (x0 ) + g + y ,

(2.3)

if x∗1 ( f (λ0 x0 )) = 0, then just take

η1 = 1 and choose η2 , ξ2 and ξ3 as uniquely determined complex numbers in S C satisfying (2.3). Notice that (λ0 x0 , λ0 z∗ ) ∈ Π( X ). Hence v (h) ϕ (x0 ) + | y ∗ (x0 )| + g + y . For any (x, x∗ ) ∈ Π( X ), ∗

x h(x) ϕ (x) + y ∗ (x) + g + y ϕ (x0 ) + y ∗ (x0 ) + g + y .

Hence v (h) = ϕ (x0 ) + | y ∗ (x0 )| + g + y . We claim that h is a numerical strong peak function at (λ0 x0 , λ0 z∗ ). Indeed, if there is a sequence ( w n , w n∗ ) ∈ Π( X ) such that limn | w n∗ (h( w n ))| = v (h), then there is a sequence {τn } in S C such that

∗

w h( w n ) w ∗ f ( w n ) + y ∗ ( w n ) + w ∗ ( y ) + g x∗ ( w n ) n n n 1

ϕ (τn w n ) + Re y ∗ (τn w n ) + w n∗ ( y ) + g x∗1 ( w n ) ϕ (x0 ) + y ∗ (x0 ) + g + y = v (h).

Therefore limn τn w n = x0 and limn x∗1 ( w n ) = λ0 . So it is easy to see that limn τn = λ0 . This implies that limn w n = λ0 x0 . Let x∗ be a weak-∗ limit point of {xn∗ }n∞=1 . Then

v (h) = x∗ h(λ0 x0 )

= x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 )x∗ (x0 ) + η2 g x∗ (x0 ) + x∗ ( y )

x∗ f (λ0 x0 ) + y ∗ (x0 )x∗ (x0 ) + g x∗ (x0 ) + x∗ ( y ) ϕ (x0 ) + y ∗ (x0 ) + g + y = v (h)

shows that |x∗ (x0 )| = 1 and |x∗ ( y )| = y . So x∗ = ξ1 x∗1 + ξ3 x∗3 + x∗4 for some x∗4 ∈ B ∞ with supp(x∗4 ) = [supp(x0 ) ∪ supp( y )]c and ξ1 , ξ3 ∈ S C . So

v (h) = x∗ h(λ0 x0 )

= ξ1 x∗1 f (λ0 x0 ) + x∗4 f (λ0 x0 ) + ξ1 η1 y ∗ (λ0 x0 ) + η2 g ξ1 + ξ3 y

= x∗1 f (λ0 x0 ) + ξ1 x∗4 f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ1 ξ3 y

x∗1 f (λ0 x0 ) + x∗4 f (λ0 x0 ) + y ∗ (x0 ) + g + y

x∗1 f (λ0 x0 ) + x∗2 f (λ0 x0 ) + y ∗ (x0 ) + g + y = v (h)

(2.4)

(2.5)

446


and |x∗4 ( f (λ0 x0 ))| = |x∗2 ( f (λ0 x0 ))|. Notice that supp(x∗4 ) = supp(x∗2 ) and

∗

x f (λ0 x0 ) =

∗ e , f (λ0 x0 ) . i

2

i ∈supp(x∗2 )

So x∗4 = ξ2 x∗2 for some ξ2 ∈ S C . By (2.3) and (2.5),

v (h) = x∗1 f (λ0 x0 ) + x∗2 f (λ0 x0 ) + y ∗ (x0 ) + g + y

= x∗1 f (λ0 x0 ) + ξ1 ξ2 x∗2 f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ1 ξ3 y

= x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ2 x∗ f (λ0 x0 ) + ξ3 y . 1

2

= λ0 ξ2 and ξ3 numerical strong peak function at (λ0 x0 , λ0 z∗ ). Since x∗ (x0 ) = λ0 , we get

ξ

1 = λ0 , ξ2

= λ0 ξ3 . Hence x∗ = λ0 z∗ and {xn∗ } converges weak-∗ to λ0 z∗ . Thus h is a

∗ In the second case, we assume that i∈ / supp(x0 ) |e i , f (λ0 x0 )| = 0 and N = supp(x0 ). Then supp( f (λ0 x0 )) ⊂ supp(x0 ). Choose a peak function g ∈ A u ( B C ) at λ0 with g and y ∈ 1 such that supp( y ) is the complement of supp(x0 ) and y . Then there is a unique element x∗3 ∈ S ∞ such that supp(x∗3 ) = supp( y ) and x∗3 ( y ) = y . Define a function h ∈ A ( B X : X ) as

h(x) := f (x) + η1 y ∗ (x)x0 + η2 g x∗1 (x) x0 + y , and define z∗

:=

x∗

such that

∀x ∈ B X ,

∗ ∗ 1 + ξ3 x3 , where if x1 ( f (λ0 x0 )) = 0, then

η1 , η2 and ξ3 are uniquely determined complex numbers in S C

∗

z h(λ0 x0 ) = x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ3 y 1

= x∗1 f (λ0 x0 ) + y ∗ (λ0 x0 ) + g + y = ϕ (x0 ) + y ∗ (x0 ) + g + y ,

(2.6)

if x∗1 ( f (λ0 x0 )) = 0, then take η1 = 1 and choose η2 and ξ3 in S C as uniquely defined numbers satisfying (2.6). Notice that (λ0 x0 , λ0 z∗ ) ∈ Π( X ). Hence v (h) ϕ (x0 ) + | y ∗ (x0 )| + g + y . For any (x, x∗ ) ∈ Π( X ),

∗

x h(x) ϕ (x) + y ∗ (x) + g + y ϕ (x0 ) + y ∗ (x0 ) + g + y .

Hence v (h) = ϕ (x0 ) + | y ∗ (x0 )| + g + y . We claim that h is a numerical strong peak function at (λ0 x0 , λ0 z∗ ). Indeed, if there is a sequence ( w n , w n∗ ) ∈ Π( X ) such that limn | w n∗ (h( w n ))| = v (h), then there is a sequence {τn } in S C such that

∗

w h( w n ) w ∗ f ( w n ) + y ∗ ( w n ) + w ∗ ( y ) + g x∗ ( w n ) n n n 1

ϕ (τn w n ) + Re y ∗ (τn w n ) + w n∗ ( y ) + g x∗1 ( w n ) ϕ (x0 ) + y ∗ (x0 ) + g + y = v (h).

Therefore limn τn w n = x0 and limn x∗1 ( w n ) = λ0 . So it is easy to see that limn τn = λ0 . This implies that limn xn = λ0 x0 . Let x∗ be a weak-∗ limit point of {xn∗ }n∞=1 . Then

v (h) = x∗ h(λ0 )

= x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 )x∗ (x0 ) + η2 g x∗ (x0 ) + x∗ ( y )

x∗ f (λ0 x0 ) + y ∗ (x0 )x∗ (x0 ) + g x∗ (x0 ) + x∗ ( y ) ϕ (x0 ) + y ∗ (x0 ) + g + y = v (h)

(2.7)

(2.8)

shows that |x∗ (x0 )| = 1 and |x∗ ( y )| = y . So x∗ = ξ1 x∗1 + ξ3 x∗3 for some x∗3 ∈ B ∞ and ξ1 , ξ3 ∈ S C . By (2.6) and (2.8),

v (h) = ϕ (x0 ) + y ∗ (x0 ) + g + y

= x∗1 f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ1 ξ3 y

= x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g + ξ3 y . 1

ξ1 = λ0 and ξ3 = λ0 ξ3 . Hence x∗ = λ0 z∗ and {xn∗ } converges weak-∗ to λ0 z∗ . Thus h is a numerical strong peak function at (λ0 x0 , λ0 z∗ ). In the last case N = supp(x0 ), let

Since x∗ (x0 ) = λ0 , we get

h(x) = f (x) + η1 y ∗ (x)x0 + η2 g x∗0 (x) x0 ,


where

η1 , η2 are uniquely determined complex numbers in S C such that ∗

z h(λ0 x0 ) = x∗ f (λ0 x0 ) + η1 y ∗ (λ0 x0 ) + η2 g 0

= x∗0 f (λ0 x0 ) + y ∗ (λ0 x0 ) + g = ϕ (x0 ) + y ∗ (x0 ) + g

447

(2.9)

if x∗0 ( f (λ0 x0 )) = 0. On the other hand, if x∗0 ( f (λ0 x0 )) = 0, then take η1 = 1 and choose η2 in S C as uniquely defined numbers satisfying (2.9). The proof that h is a numerical strong peak function at (λ0 x0 , λ0 x∗0 ) is similar to the previous ones. In either case, f − h 3 . This completes the proof. 2 Corollary 2.16. Let X = 1 . Then the set of all numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). In particular, the intersection of both the set of all strong peak functions in A ( B X : X ) and the set of all numerical strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). Proof. Since 1 has the Radon–Nikodým property, the set of all strong peak functions in A ( B X : X ) is dense [10]. Hence by Proposition 2.1, the set of all strong peak functions in A ( B X : X ) is a dense G δ -subset of A ( B X : X ). By Corollary 2.3, the set of all numerical strong peak functions is a G δ -subset of A ( B X : X ) and dense by Theorem 2.15. So we get the desired result. 2 2.3. Numerical Shilov boundary The following proposition is a numerical version of Bishop’s theorem [10]. Proposition 2.17. Let A be a subspace of C b ( B X : X ). Suppose that the set of all numerical strong peak functions in A is dense in A. Then the τ -closure of the set of all numerical strong peak points for A is the numerical Shilov boundary of A. Proof. Let

Γ :=

z, z∗ ∈ Π( X ): z, z∗ is a numerical strong peak point for A .

Notice that every τ -closed numerical boundary of A contains Γ . Hence the numerical Shilov boundary of A contains all points of τ -closure of Γ . For the reverse inclusion, let f ∈ A be fixed. Then there exists a sequence { f n }n∞=1 of numerical strong peak functions in A such that f n − f → 0. Hence | v ( f n ) − v ( f )| → 0. Let (xn , xn∗ )n∞=1 be a sequence of numerical strong peak points in Π( X ) such that v ( f n ) = |xn∗ ( f n (xn ))| for every n. Then v ( f ) = limn→∞ |xn∗ ( f n (xn ))|. Note that for every n,

∗

x f n (xn ) − x∗ ( f n − f )(xn ) x∗ f (xn ) n n n

xn∗ f n (xn ) + xn∗ ( f n − f )(xn ) , which shows that v ( f ) = limn→∞ |xn∗ ( f (xn ))| = sup{| z∗ ( f ( z))|: ( z, z∗ ) ∈ Γ }. So Γ is a numerical boundary for A. Thus the numerical Shilov boundary of A is contained in the τ -closure of Γ . 2 Proposition 2.18. Let A be a subspace of C b ( B X : X ) which contains all functions of the forms:

x∗ ⊗ y ,

1 ⊗ z,

∀x∗ ∈ X ∗ , ∀ y , z ∈ X .

Suppose that X is smooth and locally uniformly convex. Then Π( X ) is the set of all numerical strong peak points for A and Π( X ) is the numerical Shilov boundary of A. Proof. Let (x0 , x∗0 ) ∈ Π( X ). Then g (x) = (x∗0 (x) + 1)/2 is a strong peak function at x0 . Let h(x) := g (x)x0 ∈ A. Then h is a numerical strong peak function at (x0 , x∗0 ). So Π( X ) is the set of all numerical strong peak points for A. The second assertion follows from Theorem 2.19, Proposition 3.1, and the first assertion. 2 3. Negative results for denseness of numerical peak holomorphic functions It is observed in [8] that there is no strong peak function in A wu ( B L 1 [0,1] ). The following shows that there is no numerical strong peak function in A wu ( B L 1 [0,1] : L 1 [0, 1]). Proposition 3.1. There is no numerical strong peak function in A wu ( B L 1 [0,1] : L 1 [0, 1]).

448


Proof. Let {rn }n∞=1 be Rademacher functions on [0, 1]. We shall use the following basic fact observed in [18]:

lim x + rn x1 = x1 ,

∀x ∈ L 1 [0, 1].

n→∞

Notice also that if we let xn = (1 + rn )x for n 1 and for some x ∈ X , then xn weakly converges to x. Suppose that there is a numerical strong peak function f ∈ A wu ( B L 1 [0,1] : L 1 [0, 1]) at (x, x∗ ) ∈ Π( L 1 [0, 1]). Then

1 1=

x(t )x∗ (t ) dt =

1

0

x(t ) dt

0

which shows that

x∗ (t ) = sign x(t ),

a.e. t ∈ supp(x).

If we take yn = ((11++rrn))xx for each n 1, then { yn }n∞=1 weakly converges to x and n 1 ∗

x ( yn ) =

= =

1

1

(1 + rn )x1 1

(1 + rn )x1

0

(1 + rn )x1

x∗ (t ) 1 + rn (t ) x(t ) dt

supp x

1

x∗ (t ) 1 + rn (t ) x(t ) dt

1 + rn (t ) x(t ) dt = 1.

supp x

Therefore ( yn , x∗ ) ∈ Π( X ). As {|x∗ ( f ( yn ))|}n∞=1 converges to |x∗ ( f (x))| = v ( f ) and f is a numerical strong peak function at (x, x∗ ), we get that ( yn , x∗ ) τ -converges to (x, x∗ ). However

x − yn 1 x − (1 + rn )x1 − yn − (1 + rn )x1

which shows that lim infn x − yn 1 1, a contradiction.

2

Proposition 3.2. Let K be an infinite compact Hausdorff space. Then there are no numerical strong peak functions in A b ( B C ( K ) : C ( K )). Proof. It follows from the proof of Theorem 3.2(2) in [15].

2

Note that Theorem 3.2(1) in [15] implies that there exist infinitely many numerical peak functions in A b ( B C ( K ) : C ( K )). Proposition 3.3. Let K be a compact Hausdorff space and X = C ( K ). Suppose that f ∈ A b ( B X : X ) is a numerical peak function at (x0 , x∗0 ) in Π( X ). Then x∗0 = sign(x0 (t 0 ))δt0 for some t 0 ∈ K and f (x0 ) ∈ C ( K ) is a peak function at t 0 . Proof. Notice that v ( g ) = g for every g ∈ A b ( B X : X ) [2, Theorem 2.8]. Because f is a numerical peak function at (x0 , x∗0 ) we have v ( f ) = f = |x∗0 f (x0 )|. Since x∗0 1, we have x∗0 f (x0 ) = f (x0 ) = |δt0 f (x0 )| for some t 0 ∈ K . We claim that if f (x0 ) = |δt f (x0 )|, then |x0 (t )| = 1. Otherwise we have |x0 (t )| < 1. Choose y ∈ B X such that y (t ) = 1 and define a function

ϕ (λ) = δt f x0 + λ y 1 − |x0 | . Then ϕ is a holomorphic on the open unit disc in the complex plane and continuous on the closed unit disc. Notice also that |ϕ (0)| = |δt f (x0 )| = f . By the maximum modulus theorem, ϕ is a constant. Choose λ0 in S C satisfying |x0 (t ) + λ0 (1 − |x0 (t )|)| = 1. Then ϕ (λ0 ) = |δt f (x0 + λ0 (1 − |x0 |))| = ϕ (0) = v ( f ). Since (x0 + λ0 (1 − |x0 |), sign(x0 + λ0 (1 − |x0 |))δt ) is in Π( X ), we get x∗0 = sign(x0 + λ0 (1 − |x0 |))δt . Now 1 = |x∗0 (x0 )| = |x0 (t )|. This is a contradiction. Notice that (x0 , sign(x0 (t 0 ))δt0 ) is in Π( X ) and v ( f ) = |δt0 f (x0 )| shows that x∗0 = sign(x0 (t 0 ))δt0 since f is a numerical

peak function. Finally, if there is s ∈ K such that f (x0 ) = |δs f (x0 )|, then by claim, (x0 , sign x0 (s)δs ) ∈ Π( X ) and |δs f (x0 )| = v ( f ). So we get t = s. Hence f (x0 ) is a peak function at t 0 . 2

Remark 3.4. E. Bishop showed [4] that there is a compact Hausdorff space K such that C ( K ) has no peak functions. In that case, there is no numerical peak function in A b (C ( K ) : C ( K )).


449

Recall that an element x in the unit ball B X of a complex Banach space X is said to be a complex extreme point if whenever sup0θ2π x + e i θ y 1 for some y ∈ X , we get y = 0. It is easy to see that every extreme point of B X is a complex extreme point of B X . It is observed by J. Globevnik [13] that if f ∈ A b ( B X ) is a strong peak function at x0 , then x0 is a complex extreme point of B X . Otherwise, there is a nonzero w ∈ X such that x0 + λ w 1 for every λ ∈ B C . Hence the function ϕ (λ) = f (x0 +λ w ) is holomorphic on the interior of B C and continuous on B C . Notice also that ϕ = |ϕ (0)| = f . By the strong maximum modulus theorem, ϕ is constant on B C . Hence f = | f (x0 )| = |ϕ (0)| = |ϕ (1)| = | f (x0 + w )|. So | f (x0 )| = | f (x0 + w )| and x0 = x0 + w because f is a peak function at x0 . Hence w = 0, which is a contradiction to the assumption w = 0. We denote by extC ( B X ) the set of all complex extreme points of B X and by ext( B X ) the set of all extreme points of B X . Notice that an element f ∈ ext( B C ( K ) ) on a compact Hausdorff space K and only if | f | = 1 on K . So it is easy to check that f ∈ B C ( K ) is an extreme point of B C ( K ) if and only if f is a complex extreme point of B C ( K ) . Theorem 3.5. Let K be a compact Hausdorff space. Then f is a numerical peak function in A b ( B C ( K ) : C ( K )) if and only if there exist unique x0 ∈ ext( B C ( K ) ) and t 0 ∈ K such that (a) v ( f ) = f = f (x0 ) > f (x) for every x ∈ B C ( K ) with x = x0 ; (b) v ( f ) = f = |δt0 ( f (x0 ))| > |δt ( f (x0 ))| for every t ∈ K with t = t 0 . Proof. Let X = C ( K ). Notice that v ( g ) = g for every g ∈ A b ( B X : X ) [2, Theorem 2.8]. (⇒): Suppose that f is a numerical peak function for A b ( B X : X ). Then there exists a unique (x0 , x∗0 ) ∈ Π( X ) such that

∗

x f (x0 ) = v ( f ) = f . 0

Claim: f is a norm peak function at x0 . Suppose that f (a) = f for some a ∈ B X . Then there is s ∈ K such that | f (a)(s)| = f . We shall show that |a(s)| = 1. Suppose on the contrary that |a(s)| < 1. In case that s is an isolated point, consider the function ϕ : B C → C defined by ϕ (z) = f (a − a(s)χ{s} + zχ{s} )(s), where χ A is a characteristic function on A. Then ϕ ∈ A ( B C ) and attains its maximum at the interior point a(s) of B C . So the maximum modulus theorem shows that ϕ is a constant function. Notice that

f = ϕ (1) = f a − a(s)χ{s} + 1χ{s} (s)

= f a − a(s)χ{s} − 1χ{s} (s) = ϕ (−1). This means that

f = δs f a − a(s)χ{s} + χ{s} = −δs f a − a(s)χ{s} − 1χ{s} .

Since (a − a(s)χ{s} + χ{s} , δs ) and (a − a(s)χ{s} − χ{s} , −δs ) are two different elements in Π( X ), it is a contradiction to that f is a numerical peak function. For the other case, suppose that s is not an isolated point. So the set A := {t ∈ K : |a(t )| < 1} contains at least one different point other than s. So we can choose two ϕ , ψ ∈ C ( K ) such that ϕ (s) = ψ(s) = 1, ϕ | A = ψ| A and |ϕ | 1, |ψ| 1. Now choose a complex number z0 ∈ S C such that |a(s) + z0 (1 − |a(s)|)| = 1 and let w 1 := a(s) + z0 (1 − |a(s)|). Then if we consider the function z → f (a(·) + z0 ϕ (·)(1 − |a(·)|))(s), it belongs to A ( B C ) and attains its maximum f at 0. Hence it is a constant function. So f = | w 1 δs ( f (a(·) + z0 ϕ (·)(1 − |a(·)|)))| and (a(·) + z0 ϕ (·)(1 − |a(·)|), w 1 δs ) is in Π( X ). Hence x0 = a(·) + z0 ϕ (·)(1 − |a(·)|) because f is a numerical peak function at (x0 , x∗0 ). Similarly, we have x0 = a(·) + z0 ψ(·)(1 − |a(·)|). Then ϕ (t ) = ψ(t ) if |a(t )| < 1. This is a contradiction to that ϕ | A = ψ| A . Therefore we show that |a(s)| = 1. Then (a, a(s)δs ) ∈ Π( X ) and f = | f (a)(s)| shows that a = x0 . This proves that f is a peak function at x0 . Then x0 is a complex extreme point of B X . Thus x0 ∈ ext( B X ) = extC ( B X ). Since f (x0 ) ∈ C ( K ), there is t 0 ∈ K such that

sign x0 (t 0 ) δt f (x0 ) = f (x0 )(t 0 ) = v ( f ) = f = f (x0 ). 0

Claim: v ( f ) = f = |δt0 ( f (x0 ))| > |δt ( f (x0 ))| for every t ∈ K with t = t 0 . Since x0 ∈ ext( B X ), |x0 (s)| = 1 for every s ∈ K . Hence if t ∈ K and t = t 0 , then (x0 , sign(x0 (t ))δt ) ∈ Π( X ). Notice that f is a numerical peak function at (x0 , sign(x0 (t 0 ))δt0 ). Hence

v ( f ) = f = δt0 f (x0 ) > δt f (x0 ) because δt = δt0 by the Urysohn lemma. (⇐): Let x0 ∈ ext( B X ) and t 0 ∈ K satisfying the conditions (a) and (b). Claim: f is a numerical peak function at (x0 , sign(x0 (t 0 ))δt0 ).

450


Let ( y , y ∗ ) ∈ Π( X ) be such that | y ∗ ( f ( y ))| = v ( f ) = f . Since f = f ( y ) | y ∗ ( f ( y ))| = v ( f ) = f , by condition (a), we have y = x0 . By the Riesz representation theorem, there exists a unique regular complex Borel measure μ on K which represents y ∗ . So we get

f (x0 ) = y ∗ f (x0 ) =

f (x0 ) dμ. K

We will show that the measure

μ is supported only on the point {t 0 }. Let T be a compact subset of K \ {t 0 }. Because

y ∗ = 1, the total variation |μ|( K ) of μ is one and we have

f (x0 ) = f (x0 ) dμ f (x0 ) d|μ| K

K

f (x0 ) d|μ| +

f (x0 ) d|μ|

K \T

T

max f (x0 )(t ): t ∈ T · |μ|( T ) + f (x0 ) · |μ|( K \ T ) f (x0 ). Then the condition (b) means that max{| f (x0 )|(t ): t ∈ T } < | f (x0 )(t 0 )| = f (x0 ) and |μ( T )| = 0. By the regularity of μ, we get μ( K \ {t 0 }) = 0 and μ = λδt0 for some complex number λ with |λ| = 1. Because (x0 , y ∗ ) ∈ Π( X ), λ = sign(x0 (t 0 )) and we have shown that f is a numerical peak function at (x0 , sign(x0 (t 0 ))δt0 ). 2 Remark 3.6. Let K be a completely regular Hausdorff space. Then C b ( K ) is isometrically isomorphic with C (β K ), where β K ˘ is the Stone–Cech compactification of K . Therefore f is a numerical peak function in A b (C b ( K ) : C b ( K )) if and only if there exist unique x0 ∈ ext( B C b ( K ) ) and x∗0 ∈ ext( B C b ( K )∗ ) such that (a) v ( f ) = f = f (x0 ) > f (x) for every x ∈ B C b ( K ) with x = x0 ; (b) v ( f ) = f = |x∗0 ( f (x0 ))| > | y ∗ ( f (x0 ))| for every y ∗ ∈ ext B C b ( K )∗ with y ∗ = x∗0 . Remark 3.7. In general, it is not true that if f is a peak function for A b ( B C ( K ) : C ( K )), then f is a numerical peak function. Indeed, suppose that K is finite with more than two elements. Then C ( K ) = n∞ for some n > 1. Hence there is a peak function h ∈ A u ( B C ( K ) ) at x0 with h = 1 since C ( K ) is finite dimensional. Given two distinct points t 0 and t 1 in K , choose a function g ∈ C ( K ) such that g = 1 = g (t 0 ) = g (t 1 ). Hence x0 ∈ ext( B C ( K ) ) and |x0 | = 1 on K . Hence if we define f : B C ( K ) → C ( K ) by f (x) := h(x) g for each x ∈ B C ( K ) , then f = f (x0 ) = |h(x0 )| > |h(x)| = f (x) for any x ∈ B C ( K ) with x = x0 . Hence f is a peak function on A u ( B C ( K ) : C ( K )). However (x0 , sign(x0 (t 0 ))δt0 ) and (x0 , sign(x0 (t 1 ))δt1 ) are in Π( X ) and it is clear that

δt f (x0 ) = 1 = δt f (x0 ) . 0 1

Therefore f is not a numerical peak function. The holomorphic numerical index na ( X ) [17] is defined by

na ( X ) = inf v ( f ): f = 1, f ∈ A u ( B X : X ) . When X is a complex Banach space with na ( X ) = 1, we get the following. Proposition 3.8. Suppose that X is finite dimensional and na ( X ) = 1. Then f is a numerical peak function in A u ( B X : X ) if and only if there exist x0 ∈ extC ( B X ) and x∗0 ∈ ext( B X ∗ ) such that (a) v ( f ) = f = f (x0 ) > f (x) for every x ∈ extC ( B X ) with x = x0 ; (b) v ( f ) = |x∗0 ( f (x0 ))| > | y ∗ ( f (x0 ))| for every y ∗ ∈ B X ∗ with y ∗ = x∗0 and y ∗ (x0 ) = 1; (c) v ( f ) > |x∗0 ( f ( y ))| for every y ∈ B X with y = x0 and |x∗0 ( y )| = 1. Proof. Suppose that f is a numerical peak function in A u ( B X : X ) at (x0 , x∗0 ). So f = v ( f ) = |x∗0 ( f (x0 ))| = α x∗0 ( f (x0 )) for some α ∈ S C . Then the set T = {x∗ ∈ B X ∗ : x∗ (x0 ) = 1, α x∗ ( f (x0 )) = f } is a nonempty weak-∗ compact subset of B X ∗ . Hence T has an extreme point y ∗ . Since T is an extremal subset of B X ∗ , y ∗ is an extreme point of B X ∗ . Let ϕ (x) := y ∗ ( f (x)) be the function in A u ( B X ). Then by the Bishop theorem [4],

ϕ =

max

ϕ (x),

x∈ρ A u ( B X )


451

where ρ A u ( B X ) is the set of all peak points of A u ( B X ). So there is x1 ∈ ρ A u ( B X ) such that f = ϕ = |ϕ (x1 )| = | y ∗ ( f (x1 ))|. Hence x1 ∈ extC ( B X ). Since na ( X ) = 1, | y ∗ (x1 )| = 1 by Corollary 2.10 in [17]. Notice that | y ∗ ( f (x1 ))| = f = v ( f ). Since (x1 , sign( y ∗ (x1 )) y ∗ ) ∈ Π( X ) and f is a numerical peak function, we get x1 = x0 and x∗0 = sign( y ∗ (x1 )) y ∗ . Hence x0 ∈ extC ( B X ) and x∗0 ∈ ext( B X ∗ ). Since f is a numerical peak function at (x0 , x∗0 ), both (b) and (c) hold clearly. Fix x1 ∈ extC ( B X ) with x1 = x0 . Then the set S = {x∗ ∈ B X ∗ : x∗ f (x1 ) = f (x1 )} is a nonempty weak-∗ compact subset of B X ∗ and an extremal subset of B X ∗ . Hence there is y ∗ ∈ ext( B X ∗ ) such that y ∗ ( f (x1 )) = f (x1 ). Hence (x1 , sign( y ∗ (x1 )) y ∗ ) ∈ Π( X ) and f = f (x0 ) = |x∗0 ( f (x0 ))| = v ( f ) > | y ∗ ( f (x1 ))| = f (x1 ) because f is a numerical peak function at (x0 , x∗0 ). This shows that (a) holds. Conversely, suppose that f in A u ( B X : X ) satisfies both (a) and (b). By Corollary 2.10 in [17], (x0 , x∗0 (x0 )x∗0 ) ∈ Π( X ) and v ( f ) = |x∗0 ( f (x0 ))|. By (b), x∗0 (x0 ) = 1. Suppose that ( w , w ∗ ) ∈ Π( X ) such that v ( f ) = | w ∗ ( f ( w ))|. Choose γ ∈ S C such that | w ∗ ( f ( w ))| = γ w ∗ ( f ( w )). Then v ( f ) = f = f ( w ). So if we let

R = x∗ ∈ B X ∗ :

γ x∗ f ( w ) = f ( w ), x∗ ( w ) = 1 ,

then R is a nonempty weak-∗ compact subset of B X ∗ and an extremal subset of B X ∗ . Hence there is t ∗ ∈ ext( B X ∗ ) such that γ t ∗ ( f ( w )) = f ( w ) and t ∗ ( w ) = 1. We consider the function ψ ∈ A u ( B X ) given by ψ(x) = t ∗ ( f (x)) on B X . By the Bishop theorem again, there is t ∈ extC ( B X ) such that ψ = |ψ(t )| = |t ∗ ( f (t ))| |ψ( w )| = f ( w ). So f (x0 ) = v ( f ) = f ( w ) f (t ). Hence t = x0 by (a). Then |t ∗ ( f (x0 ))| = f . By Corollary 2.10 in [17], |t ∗ (x0 )| = 1. So (x0 , sign(t ∗ (x0 ))t ∗ ) ∈ Π( X ). Then (b) shows that sign(t ∗ (x0 ))t ∗ = x∗0 . Both |x∗0 ( w )| = 1 and (c) imply that w = x0 . Then by (b), w ∗ = t ∗ = x∗ . This shows that f is a numerical peak function at (x0 , x∗0 ). 2 Proposition 3.9. Let Ω be a locally compact Hausdorff space with more than 2 elements. Then there are no numerical peak functions in A b ( B C 0 (Ω) : C 0 (Ω)). Proof. Otherwise, there exists a numerical peak function f ∈ A b ( B C 0 (Ω) : C 0 (Ω)). There exists (x0 , x∗0 ) ∈ Π(C 0 (Ω)) such that

v ( f ) = x∗0 f (x0 ) > z∗ f ( z)

for every z, z∗ ∈ Π C 0 (Ω) \

x0 , x∗0

.

Note that v ( f ) = f . There exists t 0 ∈ Ω such that |δt0 ( f (x0 ))| = f (x0 ). Claim: |x0 (t 0 )| = 1 and x∗0 = sign(x0 (t 0 ))δt0 . Assume that |x0 (t 0 )| < 1. Let y 0 ∈ B C ( K ) such that y 0 (t 0 ) = 1. Define

ψ0 (λ) = δt0 f x0 + λ y 0 1 − |x0 | λ ∈ C, |λ| 1 , which is a continuous function on the closed unit disk and holomorphic on the open unit disk. By the maximum modulus theorem, ψ0 ≡ ψ0 (0) on the open unit disk. Choose λ0 ∈ C such that |λ0 | = 1 and |x0 (t 0 ) + λ0 (1 − |x0 (t 0 )|)| = 1. Let z0 := x0 + λ0 y 0 (1 − |x0 |) ∈ B C 0 (Ω) . Note that ( z0 , sign( z0 (t 0 ))δt0 ) ∈ Π(C 0 (Ω)) and

sign z0 (t 0 ) δt f ( z0 ) = ψ0 (λ0 ) = v ( f ). 0 We must have z0 = x0 , which is impossible. Thus |x0 (t 0 )| = 1 and x∗0 = sign(x0 (t 0 ))δt0 . Since x0 ∈ C 0 (Ω), there exists t 1 ∈ Ω such that |x0 (t 1 )| < 1. Clearly t 0 = t 1 . Let y 1 ∈ B C 0 (Ω) such that y 1 (t 1 ) = 1. Define

ψ1 (λ) = δt0 f x0 + λ y 1 1 − |x0 | λ ∈ C, |λ| 1 , which is a continuous function on the closed unit disk and holomorphic on the open unit disk. By the maximum modulus theorem, ψ1 ≡ ψ1 (0) on the open unit disk. Choose λ1 ∈ C such that |λ1 | = 1 and |x0 (t 1 ) + λ1 (1 − |x0 (t 1 )|)| = 1. Let z1 := x0 + λ1 y 1 (1 − |x0 |) ∈ B C 0 (Ω) . Note that ( z1 , sign( z1 (t 0 ))δt0 ) ∈ Π(C 0 (Ω)) and

sign z1 (t 0 ) δt f ( z0 ) = ψ1 (λ1 ) = v ( f ). 0 We must have z1 = x0 , which is a contradiction because z1 (t 1 ) = x0 (t 1 ).

2

Acknowledgment The authors thank an anonymous referee whose careful reading and suggestions improved the style of presentation and helped to eliminate the superfluous condition in the previous version of Theorem 3.5.

452


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